2  jf,  // 


/<?. 


1013  12th  St.,  Oakland,  Cal.  J 

Not  to  !><•  taken  from  the  Library.    ? 


THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


GIFT  OF 

John  S.Prell 


*» 

ii 


L.  M.  Clement. 


THE 


THEORY  OF  STRAINS. 


A  COMPENDIUM  FOB  THE 


CALCULATION  AND  CONSTRUCTION  OF  BRIDGES, 
ROOFS  AND  CRANES, 


WITH   THE   APPLICATION   OF 

TRIGONOMETRICAL  NOTES. 

CONTAINING 

THE   MOST   COMPREHENSIVE   INFORMATION    IN   REGARD  TO   THE 

RESULTING  STRAINS  FOR  A  PERMANENT  LOAD,  AS  ALSO  FOR 

A  COMBINED  (PERMANENT  AND  ROLLINGS  LOAD. 

IN  TWO  SECTIONS. 

ADAPTED  TO  THE  REQUIREMENTS  OF  THE  PRESENT  TIME, 

<g> 

BY 

£?,<?  4rt 

JOHN    H.   DTEDRFCHS, 

•  tt>  J*S  a.   j^-    1 


&A 

CIVIL    AND    MKCHAMCAL    ENGINEER. 


rOv 


ILLUSTRATED  BY  NUMEROUS  PLATES  AXD   DIAGRAMS 

\- 

i- 


BALTIMORE: 

SCHMIDT    &    T  E  O  W  E  . 
1871. 


PRELL 

Civil  6-  Mechanical  Engineer. 

SAN  FBAN  CISCO,  CAL. 


Entered  according  to  Act  of  Congress,  in  the  year  1871,  by 

JOHN  H.  DIEDR1CHS, 
In  tne  Office  of  the  Librarian  of  Congress,  at  "Washington. 


WESTCOTT  &  THOMSON, 
Stereotypcrs,  Philada. 


Engineering 
Library 


INSCRIBED   TO 

WENDEL    BOLLMAN,  ESQ., 

IN  TESTIMONY   OF  HIS  INTEREST  IN  SCIENCE  AND  ART. 


- 


l\ 


k 


713709 

Library 


CONTENTS. 


PREFACE .. ....... 7 

EXPLANATION  OF  CHARACTERS  USED  IN  THE  CALCXTLATIONS 11 

SECTION   I. 

A.  INTRODUCTION 13 

I.  THE  LEVER 15 

II.  SUSPENDED  WEIGHTS  AND  THE  RESULTING  STRAINS 18 

III.  TRUSSES  WITH  SINGLE  AND  EQUALLY-DISTRIBUTED  LOAD 21 

Suspension  Truss  Bridge , 23 

Suspension  Bridge 27 

B.  ROOF  CONSTRUCTION 31 

C.  SEMI-GIRDERS 43 

I.  SEMI-GIRDERS  LOADED  AT  THE  EXTREMITY... 43 

Cranes . , 44 

II.  SEMI-GIRDERS  LOADED  AT  EACH  APEX 45 

D.  GIRDERS  WITH  PARALLEL  TOP  AND  BOTTOM  FLANGES  48 

I.  STRAIN  IN  DIAGONALS  AND  VERTICALS .. 48 

II.  STRAIN  IN  FLANGES , ...» 51 

III.  TRANSFORMATIONS .. 53 

IV.  GENERAL  REMARKS...... ..... 53 

Directions  for  the  Calculation  of  Complex  Systems 54 

E.  COMPARATIVE  TABLES  OF  RESULTING  STRAINS  FOR  A  " 

PERMANENT  LOAD ., 56 

I.  SYSTEM  OF  RiGicr-AN«u;r>  TRIAVGI.KS , 50 

II.  SYSTKM  OF  ISOSTRI.KS  BHACIXO.... oS 

i  *  t> 


CONTENTS. 


SECTION   II. 

GIRDEKS  CALCULATED  FOR  COMBINED  (.PERMANENT  AND 
ROLLING)  LOAD. 

PAG* 

A.  GIRDERS  WITH  PARALLEL  TOP  AND  BOTTOM  FLANGES  59 

I.  THE  RIGHT-ANGLED  SYSTEM 5!) 

II.  ISOSCELES  BRACING 66 

1.  Triangular  Truss 66 

Calculation  of  Strains  y  and  u  in  Diagonals 67 

2.  Isometrical  Truss 69 

a.  Calculation  of  Diagonals 70 

b.  Calculation  of  Top  and  Bottom  Chords  (Flanges) 72 

B.  CAMBER  IN  TRUSSES,  WITH  PARALLEL  TOP  AND  BOT- 

TOM CHORDS 73 

TABLES   CONTAINING   THE  LENGTH  OF  ARCHES  FOR   DEGREES, 

MINUTES  AND  SECONDS,  FOR  A  RADIUS  AS  UNIT 77 

0.   PARABOLIC  GIRDER  OF  48  FEET,  OR  16  METRES,  SPAN  80 

D.  THE  ARCHED  TRUSS 86 

Calculation  of  Strain  y  in  the  Diagonals 88 

Calculation  of  Strain  in  the  Verticals  V. 90 

Transformations 93 

E.  THRUST  CONSTRUCTION 94 

I.  GIRDER  20  FEET  IN  LENGTH  (WITH  A  SINGLE  WEIGHT  AT  THE 

CENTRE). 

II.  GIRDER  72  FEET  IN  LENGTH  (CALCULATED  FOR  PERMANENT 
AND  ROLLING  LOAD). 

Definition  of  Strain  x  in  the  Horizontal  Flanges 97 

Definition  of  Strain  y  in  the  Diagonals 98 

Calculation  of  the  Tensile  Strains  z  in  the  Lower  Flanges 99 

Calculation  of  the  Verticals  u 101 

III.  CALCULATION  OF  A  TRUSS  SUSTAINING  A  DOME 102 

Calculation  of  Strains  x  of  the  Outside  Arch 104 

Calculation  of  Strains  z  of  the  Inside  Arch 10o 

Calculation  of  the  Diagonals  y 106 


PREFACE. 


THE  want  of  a  compact,  universal  and  popular  treatise  on 
the  construction  of  Hoofs  and  Bridges — especially  one  treating 
of  the  influence  of  a  variable  load — and  the  unsatisfactory 
essays  of  different  authors  on  the  subject,  induced  me  to  pre- 
pare the  following  work. 

Bridge-building  has  been  and  always  will  be  an  important 
branch  of  industry,  not  only  to  engineers,  but  also  to  the  masses 
for  the  purposes  of  travel  and  trade,  and,  as  Colonel  Merrill  in 
his  recent  essay  on  Bridge-building  remarks,  "  important  to 
railroad  companies  on  account  of  the  large  amount  of  capital 
invested  in  their  construction." 

Bridge  literature  has  often  been  used  by  rival  parties  for  the 
purpose  of  advancing  their  own  private  interests,  their  motive 
being  competition.  Imposing  upon  the  faith  and  credulity  of 
those  whom  they  pretend  to  serve,  there  is  no  guarantee  that 
worthless  structures  will  not  be  erected. 

Thoroughly  independent  of  any  such  motive,  my  aim  is  to 
give,  especially  to  bridge-builders  and  to  engineers  and  archi- 
tects, the  results  of  my  investigation  on  the  subject  of  calcu- 
lating strains,  in  order  that  capitalists  and  the  public  may  be 
benefited  and  protected. 


_J  PREFACE. 

These  calculations  will  also  unable  those  who  have  but  a 
limited  knowledge  of  mathematics  to  acquire  the  necessary 
information.  For  this  reason  special  attention  is  paid  to  the 
arrangement  of  the  work,  the  whole  being  made  as  plain  ami 
simple  as  possible,  in  order  to  meet  the  wants  of  the  common 
mechanic  as  well  as  the  experienced  engineer. 

Though  there  are  many  valuable  treatises  of  this  kind,  there 
has  as  yet  been  no  work  published  serviceable  to  the  degree 
desired  by  the  practical  builder  or  mechanic — most  of  the  dis- 
sertations being  too  theoretical  and  hard  to  comprehend  by  one 
not  versed  in  the  higher  mathematics;  and  some  are  so  ar- 
ranged that  a  clear  understanding  of  the  calculations  is  very 
difficult.* 

The  most  valuable  work  in  the  language  is  doubtless  Mr. 
Stoney's  "  Theory  of  Strains,"  though  the  Method  of  Moments 
is  not  developed  to  that  degree  which  I  think  necessary  for  the 
practical  man. 

We  owe  to  the  renowned  German  engineers  Hitter  and  Von 
Kaven  the  universal  application  of  this  Method  in  the  work 
entitled  "Dach  und  Brucken  Constructionen,"  in  which  it  is 
fully  explained  by  examples  and  illustrated  by  diagrams,  these 
being  often  carelessly  neglected  in  other  works. 

The  above-mentioned  work  served  me  very  much  in  the 
arrangement  of  this,  which  I  hope  will  be  kindly  received. 

The  work  being  expressly  prepared,  as  aforesaid,  for  the  use 
of  beginners  in  the  study  of  mathematics,  as  well  as  for  the 

*  As  an  exception,  may  be  named  Mr.  Shreve's  brief  but  popular  treatise 
in  Van  Nostrand's  "Engineering  Magazine,"  No.  3 x.,  August,  1870;  Vol.  III. 


PliEI'ACJE,  0 

more  advanced  practical  engineer,  it  will  enable  them,  after  a 
short  perusal,  to  acquire  all  the  necessary  information,  for  which 
even  the  trigonometrical  notes  accompanying  the  general  results 
are  not  really  required. 

The  higher  classes  of  colleges  and  other  institutions  of  learn- 
ing will  find  the  work  very  valuable. 

On  account  of  the  expense, 'an  intended  Appendix,  contain- 
ing a  rational  and  concise  investigation  on  "  The  Strength  of 
Materials,"  had  to  be  dispensed  with;  yet  I  hope  with  this 
volume  to  gratify  not  only  the  desire  of  friends,  but  to  be  able 
with  great  satisfaction  to  assist  engineers  in  the  pursuit  oi' 

their  high  and  noble  calling* 

THE  AUTHOR, 
OCTOBER,  1870. 


EXPLANATION   OF   CHAKACTEES    USED  IN   THE 
CALCULATIONS. 


=  Equal,  or  the  sign  of  equality. 

-f-  Plus,  or  the  sign  of  addition;  also,  the  symbol  of  positive  (tensile)  strain. 

—  Minus,  or  the  sign  of  subtraction ;  also,  the  symbol  of  negative  (compres- 

eive)  strain. 

X  or  .  Sign  of  multiplication. 
:  or  -5-  Sign  of  division. 
,   Sign  of  decimals ;  also,  of  thousands, 
oo   Sign  of  infinite. 
<  Sign  of  angle,  signified  by  the  Grecian  cyphers   oc,  /?,  y,  6f  e. 

2  Sign  of  square  of  a  number. 
1/~~  Sign  of  square  root  of  a  number. 
0    Sign  of  degrees. 
'     Sign  of  minutes;  also,  of  feet. 
ff  Sign  of  seconds;  also,  of  inches. 
(     )  or  [     ]  Brackets,  to  enclose  the  mathematical  expression  bound  to  the 

same  operation. 

TT    The  number  3,  14,  or  periphery  for  a  unit  of  the  diameter. 
E    Eight  angle,  or  90°. 
J_  Vertical. 

II 


THE  THEORY  OF  STRAINS. 


SECTION  I. 

A.    INTRODUCTION. 

To  enable  the  student  to  comprehend  the  work,  and  have  a 
thorough  knowledge  of  certain  conditions  and  examples  without 
studying  the  whole,  it  is  necessary  for  him  to  understand  the 
arrangement  of  the  following  pages. 

On  the  first  few  pages  and  the  appertaining  figures  at  the  close 
of  the  chapter  is  found  a  short  description  of  the  lever  in  its  differ- 
ent appliances,  the  application  being  only  a  key  to  the  calculations 
of  strains  which  follow. 

The  trigonometrical  notes  are  in  many  cases  almost  superfluous. 
Still,  it  may  be  advantageous  in  this  way  to  accustom  the  reader  to 
their  use.  The  "  Suspended  Weights  and  Resulting  Strains"  are 
developed  by  the  parallelogram  of  forces,  and  for  a  plain  illustra- 
tion the  results  are  appended  to  the  figures,  which  will  also  be 
observed  on  figures  of  "Trusses  with  Single  and  Distributed  Load." 

In  the  "Suspended  Weights  and  Resulting  Strains"  a  more 
elaborate  calculation  was  thought  necessary,  and  therefore  an  Intro- 
duction to  the  calculation  by  the  "  Method  of  Moments"  may  be 
found  in  its  proper  place. 

This  Introduction  presents  the  beginner  with  a  clear  and  com- 
prehensive knowledge  of  the  formation  of  Moments;  and  the  equa- 
tions for  Figs.  20,  21,  etc.,  explain  the  equilibrium  of  force  and 
leverage. 

At  the  close  of  this  chapter  is  found  the  explanation  of  maxi-v 
mum  compressive  and  tensile  strain  in  the  top  and  bottom  chords 
of  a  parallel-flanged  truss  or  girder. 

On  "Roof  Construction"  (B),  Plates  6  to  11,  remarks  are  u-i- 

2  13 


14  THE   THEORY    OF   STRAINS. 

necessary.  The  builder  can  with  ease  find  from  the  figures  a 
system  to  suit  his  purpose.  (See  also  "  Arched  Trusses,"  D,  Sec- 
tion II.)  '« 

On  "Semi-Girders"  (C)  the  calculation  of  strains  is  treated  in 
the  way  heretofore  generally  known  (determining  from  the  centre 
toward  the  abutment),  after  which  the  "Method  of  Moments"  is 
applied  to  the  same  example,  followed  by  a  more  elaborate  expla- 
nation of  the  principles  of  Moments  on  the  crane  skeleton  (Plate 
14),  whose  single  members  are  altogether  divergent. 

The  thorough  calculation  of  a  truss  with  horizontal  top  and 
bottom  flanges  (right-angled  system  D,  Sect.  I.),  with  the  resulting 
strains  for  a  system  of  braces,  all  of  which  are  inclined  ii^  the 
same  direction,  shows  how  easily  by  transformation  of  the  strains 
a  system  of  bracing  just  reversed  can  be  formed.  (Comp.  D, 
Sect,  II.) 

From  the  comparative  tables,  E,  L,  II.  (Plate  18,  19)  those  who 
are  not  mathematicians  can  find,  by  a  little  study  (for  an  assumed 
load  "  W" — a  variable  load  not  considered),  the  strains  in  flanges, 
braces  and  ties.  (See  D,  IV.,  Sect.  I.) 

The  progress  of  panels,  and  by  this  the  increase  of  stress  in  the 
different  members,  are  ad  libitum  to  be  extended  (are  optional). 

Where  no  composite  strains  appear,  in  the  skeletons  double  lines 
make  the  compressive  ( — )  strains  more  obvious,  the  tensile  (+) 
strains  being  always  represented  by  single  lines.  The  assumed 
load  in  the  calculations  is  equally  divided  on  the  apexes ;  but  in 
general  some  attention  may  be  paid  to  a  peculiar  load— say  from  a 
single  locomotive — at  a  certain  apex,  this  being  observed  in  exam- 
ples on  "Suspension  Truss"  (III.,  Sect.  I.). 

The  calculations  in  Sect.  II.  with  regard  to  the  influence  of  a 
variable  load  are  more  difficult  to  understand.  Still,  by  the 
results  of  strains  in  the  skeletons  it  js  easy  enough  to  form  an  idea 
about  this  matter,  and  to  see  the  importance  of  counter-bracing  or 
tying  at  centre  of  railroad-bridge  trusses. 

What  experience  and  observation  have  already  taught  to  the 
practical  railroad  man  is  here/w%  shown  by  figures. 

Information  is  given  on  parallel-flanged  trusses  for  the  so-called 
"  Camber  in  Bridges"  at  B,  Sect.  II.,  which  to  many  builders  has 
heretofore  been  only  a  matter  of  experiment. 

Yet  it  is  to  be  remarked  that  for  the  calculations  E,  Sect.  II.,  Plates 


29  to  34,  a  variation  in 
will  be  perceived,  the 
being  the   horizontal 
vertex  (centre),  an' 
ments. 

Farther  expl 
referred  to. 

The  calcuh 
nectioris   at 
relieving  tf~~         " 
strain.     T] 
ture  are  ;_J 


•* 


/» 


r  i  i  1 

f  ; 

p.     -f       *      -£ 

ur)      (*>-)      W 

^ 

5 

*  f 


^     ^ 


'  'STRAIN 

'necessary.      The  builder   can  with  east  the  fulcrum  equal 

system  to  suit  his  purpose,     (bee  also     A 
tion  II.) 

On  "Semi-Girders"  (C)  the  calculation  of 
the  way  heretofore  generally  known  (determine 
toward  the  abutment),  after  which  the  "  Method 
applied  to  the  same  example,  followed  by  a  more  < 
nation  of  the  principles  of  Moments  on  the  crane  s. 
14),  whose  single  members  are  altogether  divergent.       is  divided  in 

The  thorough  calculation  of  a  truss  with  horizont 
bottom  flanges  (right-angled  system  D,  Sect.  I.),  with  th 
strains  for  a  system  of  braces,  all  of  which  are  inclin 
same  direction,  shows  how  easily  by  transformation  of  th 
a  system  of  bracing  just  reversed  can  be  formed.      (Co) 
Sect,  II.) 

From  the  comparative  tables,  E,  I.,  II.  (Plate  18,  19)  thosi 
are  not  mathematicians  can  find,  by  a  little  study  (for  an  assu 
load  "\V" — a  variable  load  not  considered),  the  strains  in  flanpad, 
braces  and  ties.     (See  D,  IV.,  Sect.  I.) 

The  progress  of  panels,  and  by  this  the  increase  of  stress  in  thi 
different  members,  are  ad  libitum  to  be  extended  (are  optional). 

Where  no  composite  strains  appear,  in  the  skeletons  double  lines 
make  the  compressive  ( — )  strains  more  obvious,  the  tensile  (-(-) 
strains  being  always  represented  by  single  lines.  The  assumed 
load  in  the  calculations  is  equally  divided  on  the  apexes ;  but  in 
general  some  attention  may  be  paid  to  a  peculiar  load— say  from  a 
single  locomotive — at  a  certain  apex,  this  being  observed  in  exam- 
ples on  "Suspension  Truss"  (III.,  Sect.  I.). 

The  calculations  in  Sect.  II.  with  regard  to  the  influence  of  a 
variable  load  are  more  difficult  to  understand.  Still,  by  the 
results  of  strains  in  the  skeletons  it  |s  easy  enough  to  form  an  idea 
about  this  matter,  and  to  see  the  importance  of  counter-bracing  or 
tying  at  centre  of  railroad-bridge  trusses. 

What  experience  and  observation  have  already  taught  to  the 
practical  railroad  man  is  here/«M/  shown  by  figures. 

Information  is  given  on  parallel-flanged  trusses  for  the  so-called 
"  Camber  in  Bridges"  at  B,  Sect.  II.,  which  to  many  builders  has 
heretofore  been  only  a  matter  of  experiment. 

Yet  it  is  to  be  remarked  that  for  the  calculations  E,  Sect.  II.,  Plates 


p- 


a, 
*S 


/r* 

•*  7 

*                     > 

1    * 

/' 

t                                                   / 

r-               f               r                * 

^)    (?r)     (S)      '      ^ 

5 

THE     LEVER.  17 

In  every  triangle  the  sum  of  enclosed  angles, 
2  R.  =  2  X  90°, 

g-j    so  in  the  right  angled  triangle  ABC,  Fig.  6,  the  angles  A 
and  C  together  =  90°  =  R,  because  angle  B  =  90°. 

In  Trigonometry  we  say — 

b  b  c 

-  —  sine  oc ;       =  tangent  <x  ;  -  =  secant  cc. 
c  a  a 

an  c 

-  —  cosine  oc;     =  cotangent  cc;  -  —  cosecant  oc;  or, contracted, 
c  b  b 

sin.,  cos.,  tang,  or  tg.,  cotang.  or  cotg.,  sec.  and  cosec. 

For  a  radius,  AC,  as  a  unit,  the  line  AB  simply  is  called  sine; 
the  central  distance,  BC,  cosine;  and  BE  the  versed  sine  of  the 
angle  <x. 

For  certain  angles  oc  the  relation  -,-,-,  etc.,  have  certain  and 

c   a  a 

distinct  numerical  values.  (See  Haslett's  or  HaswelPs  "  Tables  of 
natural  sines,  cosines,  tangents  and  cotangents  from  1  to  90 
degrees.") 

Each  triangle,  AB.C,  A^C  (Fig.  7),  consists  of  six  members — 
i.e.,  three  sides  and  three  angles,  from  which  always  three  are  de- 
pendent on  the  rest ;  therefore,  when  three  out  of  these  six  mem- 
bers are  known,  we  can  construct,  or  with  more  exactness  we 
can  calculate,  the  others,  provided  one  at  least  of  the  given  parts 
is  a  side. 

For  the  transformation  of  trigonometrical  functions,  short  notices 
in  the  form  of  a  table,  also  the  numerical  values  (natural  sin,  cos, 
etc.)  of  the  principal  angles,  may  be  serviceable,  viz. : 

1  tang  x 


sin  x  =  - —     —  =  — - —  =  v  1  —  cos2  x. 
cosec  x         sec  x 

1  cotang  x         /- r~ 

cos  x  =  -  -  V   1 —  sin   x. 

sec  x         cosec  x 

1  sin  x 

tang  .r  =  =  V  sec''  —  1. 

cutang  x        cos  x 

cotang  x  =  -  -  =  V  cosec2  x  —  1. 

tang  .r       sin  x 

2*  R 


18 


THE 
1 

sec  x  =  — —  — 


cosec  x 


cos  x 
1 


.:Y  OF  STRAINS. 

cosec  x 

cotang  x 

sec  x 


,   -i    i    ,       2 
=  v  !  +  tauS 


tang  a? 


=  V\  -4-  cotang4  x. 


DEGREES. 

SIN. 

Cos. 

TANG. 

COTANG. 

0 

0 

1 

0 

OC 

15 

0,258 

0,965 

0,267 

3,732 

20 

0,342 

0,939 

0,363 

2,747 

25 
30 

0,422 
1=0,8 

,-      0,906 

£1/3  =  0.866 

,-      0,466 
|T/3  =  0,577 

/-      2,144 
V  3  =  1,732 

40 
45 

,_      0,642 
$1/2  =  0,707 

._      0,766 
JK  2  =  0,707 

0,839 
1,00 

1,191 
1,00 

50 

(JO 

,_      0,766 
11/8  =  0,866 

0,642 

i=o,5 

,-•_      1,191 
1/3=1,732 

._      0,839 
11/3  =  0,377 

65 

0,906 

0,422 

2,144 

0,466 

75 

0,965 

0,258 

3,732 

0,267 

12=--   90 

+  1 

0 

-f  « 

0 

2  JJ  =  180 

0 

j 

0 

—  or: 

3  11  =  270 

-••"•  1 

0                      —a: 

0 

4  /2  —  yen  i           o 

+  1 

0 

+  <* 

sin  (R  —  ye)  =  +  cos  .r,  and  cos  (7?  —  #)  =  +  sin  x. 
sin  (J2  -}-  .T)  =  +  cos  z,  and  cos  (S  -j-  a:)  =  ~  sin  x. 


II.   SUSPENDED    WEIGHTS  AND   THE  RESULTING 
STRAINS. 

In  Fig.   8",   when    IF  —  5000   Ibs,   06  =  6e  =  10, 
=  8,  and  ac  =  ce  —  12,8  feet ;  the  vertical  strain  at 


Plate  2,1 

Fig.  8*, 

"     8h. 


,        .  W      5000 

c  on  each  string  =  —  =  - — —-. 

2          2 


4000  Ibs, 


And,  further,  the  actual  strain  in  the  direction  of  the  string, 

W  ac 

P. --=9=    ir.;-» 

2     bo 

or  '=     =§009       12'8 

2  8 

All  other  information  is  given  in  Fig.  8h. 

When  a  heavy  body,  ABCD  (Fig.  8°),  is  suspended  by  two 
8,-|  oblique  strings,  DH  and  CH,  in  a  vertical  plane,  a  straight 
line  drawn  through  the  intersection  \vill  pass  through  the 
centffe.of  gravity,  G,  of  the  body. 


SUSPENDED    WEIGHTS    AND    RESULTING  STRAINS. 

For  the  force  in  the  direction  ad,  represented  by  g,  we  find 

9>]    from  Fig.  9,  „ 

od  ,     " 


i    o 
s&  ab  .  ~ 


tci  ===  0/1  +  <z  •  cos 

W  bd 


mid  in  the  same  manner  from  similarity  of  triangles, 

w  a^    — 

Iu  the  equation  for  q  is  TF.|>  the  vertical  strain  at  d  for  the 
string  ad ;  in  the  second  equation  is  F.  f '  the  vertical  strain  at 
d  for  the  string  ^-equal  to  the  shearing  strains  7  and  7,  on  the 
supports. 

Example.— When,  again, 

W  -.  5000  Ibs.,  L  —  100  feet, 
ab  '=*  10',  be  —  90',  and  bd  =  8', 
ad  =  12,84',  and  c<Z  =  90,35', 

so  p  =  5000  X  --  -  X  -^--  —  564^  lbs-» 

90        12,cS4  0(  y^>   >:^ 

n  '     1  f7    rrrz    OOUw     x\  '~    i/\ 

'  ioo        t> 

then  the  result*  for  the  horisoiitel  .train  x  and  the  vertical  strain 
Fat  the  right  support  are- 
eft      •  Trr  a6     **'. 


«  -  5000  X          X        =  5625  Ib,  ; 


20  THE   THEORY    OF   STRAINS. 

V  =  5000  X  -       =  500  Ibs. 

The  results  for  the  horizontal  strain  Xi  and  the  vertical  strain  7, 

at  the  left  support  are  — 

be    ab 

*i  =  "-T-rj' 
L    bd 

he 

and  \\  =  W.~; 

lj 

thus  x,  =  5000  X  -    ':  X  -\°  =  5625  IKs., 

100         o 

on 
and  }\  =  5000  .  —  ==  4500  Ibs.  ; 

therefore,  also,  x  =  xlt 

...  ab     be        Trr  be     ab  ,^.      1-  , 

for  W.~-.r=W.  T  •-.-    •  (Fig,  17.) 

I,     M  L     Id 

1Q  ,        For  Fig.  10,  when  IF  =  5000  Ibs.  ;   aft,  ad  and  bd  the 

same  as  before  —  • 

/if?  1  '>  .H4 

F  =  IF.  —  =  5000  X  -      -  =  8025  Ibs., 
bd  8 

and  Yl  —  IF.  f67  =  5000  X  -3°  ='  6250  Ibs. 

bd  o 

11.]     In  Fig.  12, 

P  :  Q  :  R  —  sin  .  fdb  :  sin  .  ac/6  :  sin  adc. 
12.]     In  general,  for  every  triangle, 

y  =  x-\-z,  (Fig.  12.) 

and,  as  here,  x  -f-  y  =  2  -j-  m  =  J?, 


or,  m  =  2x. 

Also,  from  similarity  of  triangles  ABC,  ADC  and  DRC, 

c        b  b> 

-  =•--  -,         ore  =  -  ; 
o        a  a 

t* 

and  as          -4.B  =  c  -f-  °»         the  diameter  =  -  -f-  a; 

a 

t.  e.,  the  diameter  equals  the  square  of  one-half  tjie  chord  divided 
by  the  height  of  the  arc  added  to  the  height  of  the  arc. 

The  height,  of  the  arc  CH  results  from  the  chord   CD  in   the 
•arne  way. 


TRUSSES  WITH  SINGLE  AND  EQUALLY-DISTRIBUTED  LOAD.      21 

Application  in  "  Camber  in  Bridges,"  B,  Sect.  II.     There,  also, 
the  geometrical  rule, 

arc  angle  at  centre 

circumference    '  360° 

[Plate  2— Figs.  8  to  12.] 


III.    TRUSSES  WITH  SINGLE  AND  EQUALL  ^-DIS- 
TRIBUTED LOAD. 

Plate  3,1         A  most  frequent  structure  is  the  trussed  benm  (Fig. 
Fig.  13J     13). 

The  post  at  the  centre  is  called  the  king-post. 

The  whole  is  a  combined  system,  in  which  the  horizontal 
beam,  according  to  its  stiffness,  relieves  the  tie-rods  from  an 
aliquot  amount  of  strain. 

For  the  greatest  exertion  to  which  the  tie-rods  in  the  most  un- 
favorable case  could  be  exposed,  we1  may  use  the  result  from 
Fig.  8b,  under  the  supposition  that  the  horizontal  beam  counteracts 
only  the  horizontal  forces.  (For  instance,  when  butted  at  the 
centre.) 

To  compute  the  stress  we  have  the  following: 

W  ac        W 

p  =  n  =  —  —  —  --  sec  .  ac, 
2    be        2 

for  a  single  weight  at  centre. 

Example.  —  The  assumed  weight  —  20000  Ibs.  ;  between,  supports, 
24  feet. 

The  length  be  =  5/>9',  and  ac  —  13,24  (which  can  be  measured 
near  enough  for  most  purposes  from  a  skeleton)  ; 

W  rir  1  ^  24- 

P  =  q  =  ~-r=:  10,000  X  ^-  =  23,700  (appror.). 
2    be  5,59 

The  angle  <x  being  in  this  case  65°,  for  a  calculation,  using  the 
preceding  tables, 


=  g=    -  sec  «  =  X  --  -  10000  X  -=  23700, 

2  2  cos  «  0,422 

be  =  ub  .  tang  25°=  12  X  0,4663  =  5,595, 
.  -  *  .  «c  .  25«  =  rf  .  =  12  X  JL.  =  13,24. 


22  THE  THEORY   OF   STRAINS. 

The  vertical  pressure  in  the  king-post  under  the  supposition  be* 
fore  =  20,000  Ibs.  ;  at  each  support  —  10,000  Ibs.  ;  and  the  conv 
preision  in  the  horizontal  beam, 

H  =  K.  fll  ==  10000  X       -  *=  21467  Ibs. 
2    be  5,59 

When  in  Fig.  15"  the  strains  p  or  q  and  the  angle  ?  are 
known,  we  find  the  resulting  vertical  strain,  E,  also  by 
means  of  the  parallelogram  of  forces,  viz., 

-^  —  Vp1  -j-  (f  -(-  2  p  .  5  .  cos  f  ; 
or,  because  in  this  case  at  =  ft,      .  therefore,  jp  =  q  ; 
also,  r  =  2.65°  =  130°,  or  =  E  +  40, 

and          cos  (R  -\-  40)  =  —  sin  40  (see  preceding  table), 


BO        R  =  1/2  p2_j_  2  ^2(—  sin  40)  =  1/2  j92  (1  —  sin  40), 
15".  J        E  =  1/2  x  237002  (1  —  0,642)  =  20000  Ibs.* 

For  a  structure,  Fig.  15b,  reserved  to  the  preceding  one  (15a),  the 
numerical  value  of  strains  is  quite  the  same,  but  of  opposite  cha- 
racter, provided  the  enclosed  angles  are  the  same* 

If,  as  in  Fig.  16,  the  load  —  40000  Ibs.,  equally  distributed  on  the 
beam,  then  each  support  will  sustain  again  one-half  of  the  load  ; 
1«-i  but  the  reaction,  D,  of  each  support  will  be  only  one-quar- 
ter of  the  load  =  10000  Ibs.  ;  and  for  the  same  exertion  a 
truss  or  beam,  charged  with  an  equally-distributed  load,  will  sustain 
twice  as  much  as  when  loaded  with  a  single  weight  at  the  centre, 
(Comp.  Fig.  14.) 

The  distribution  of  forces  on  supports  and  at  the  centre  is 
explained  by  Fig.  16*. 

For  an  angle,     <*  =  63°  26',         of,  also,  ft  =  26°  34', 
the  depth  of  the  truss  being  always  equal  to  one-fourth  of  its  length, 
for  tang  .  26°  34'  =  0,5. 

Kow,        e-  =  tang  .  ft,        or,  be  =  db  .  0,5  =  12  X  0,5  =  6, 

00 

*  For  the  angle  aed  •=  <f  in  Fig.  15h, 


R  =  yp2  +  9*  +  2  pq  cos  rf,         and  for  the  angle  edS  =  f, 
R  =  tf~Pi*  +P-  —  2  p,  .  p.  cose,     (See  Fig,  8b  and  Fig.  9.) 


^r-rs* 


SUSPENSION    TRUSS    BRIDGE,  23 

and  (Fig.  14),  the  horizontal  strain, 

ff=jri2  =  20g0012        „*=  20000, 

26  2  6 

thus  being  in  this  case  the  same  as  the  weight,  W,  at  the  centre. 

The  horizontal  strain  (thrust)  and  the  strain  on  the  oblique  rods 
increase  with  the  angle  <xj  thus  being  GO  for  an  angle  <x.  =  90°. 

SUSPENSION  TRUSS  BRIDGE. 

We  find  a  combination  of  trussing  in  the  well-known  "Suspen- 
sion Truss"  bridges,  the  principles  for  calculation  of  the  strains 
being  contained  in  the  preceding. 

The  Bollmann  truss  forms  a  continuous  system  of  independent 
trusses,  in  number  equal  to  the  number  of  vertical  posts  combined 
to  a  common  top  chord  (stretcher). 

Plate  4,1  By  Fig.  17  the  dimensions  of  such  a  truss  may  be 
Fig.  17.J  represented. 

When  for  a  single-track  railroad-bridge  the  assumed  load,  in^ 
eluding  the  weight  of  structure  —  1£  tons  =  3360  Ibs.  per  lineal 
foot,  or  for  one  rib  (single  truss)  =  1680  Ibs.  per  lineal  foot — i.  e., 
for  the  given  dimensions  12  X  1680  =  20160  Ibs.  on  each  post, 
for  which  may  be  said  in  round  figures  20000  Ibs.  =  W,  for  calcu- 
lation, then  in  Fig.  17,  according  to  Fig.  9,  the  tension  in  the  first 
rod  nearest  the  abutment, 

Strain  No.  1,  rod  =  W-  —  - Ak-  9*  20000  X  -  X  ^-=27300 Ibs., 

the  section  of  which  for  a  value  of  iron  =  10000  Ibs.  per  square 
inch  (five  to  six  times  security)  —  2,73  square  inches ;  thus,  when 
two  rods  are  applied,  the  size  of  each  rod  =  1  X  It". 

Strain  No.  2,  rod  ==  W-  —  •  ~  =  20000  X  -  X  —  =  39000  Ibs, 
AF    d  8       10 

Section  —  3,9  sq.  in.,  or  2  rods,  each  1  X  2". 

Strain  No.  3,  rod  =  W-  —  .  —  =  20000  X .-  X  ^-  =  46625  Ibs. 
AF  dm  8         10 

Sect.  =  4,66,  or  2  rods,  each  1  X  2f". 

FF     AN  4.       4Q 

Strain  No.  4,  rod  =  W-  --  =  20000  X  -  X  —  =  49000  Ibs. 

AF   EN.  8      10 

Sect,  33  4,90,  or  2  rods,  each  1  X  2i". 


24  THE    THEORY    OF    STRAINS. 


Strain  No.  5,  rod  =  W-  ~'       =  2000°  X     X     ~  =  45600  Ibs. 


Sect.  =  4,56,  or  2  rods,  each  1  X  2t". 

Strain  No.  6,  rod  =  W-  -ff~-  •  ~P  =  20000  X  f  X  "^~  =  36300  Ibs. 
yljP  gp  8        10 

Sect.  =  3,63,  or  2  rods,  each  1  X  It". 

Strain  No.  7,  rod  =  W-  —  •  —?-  =  20000  X  -  X  —  =  21150  Ibs. 
AF    hq  8        10 

Sect,  =  2,11,  or  2  rods,  each  i  X  It". 

When  for  a  partial  load  at  a  certain  panel  the  exertion  of  a  pair 
of  suspenders  is  greater  than  for  a  distributed  load  in  calculation, 
those  rods  would  be  strained  more  than  to  one-fifth  or  one-sixth  of 
their  ultimate  strength.  So,  when  a  locomotive  of  84000  Ibs. 
weight  rests  at  a  certain  panel  on  a  wheel-base  of  12  feet,  to  each 
of  the  four  supporting  posts  would  be  transmitted  one-fourth  of  its 
weight  =  21000  Ibs.  —  this  differing  very  little  from  the  calcula- 
tion in  the  example.  Additional  rods  (panel-rods)  are  applied, 
sustaining  the  main  ,  suspenders  and  at  the  same  time  the  top 
chord,  transmitting  and  distributing  the  weight  on  the  posts,  these 
being  always  in  a  state  of  compression  equal  to  the  weight  on  the 
post. 

Without  the  panel-rods  for  an  over-grade  bridge  (through  bridge) 
there  would  be  in  the  post  no  further  compression  than  that  pro- 
duced by  the  weight  of  the  top  chord  and  appendages,  leaving  for 
a  strong  cambered  truss  (B,  Sect,  II.),  in  case  of  a  partial  load,  the 
possibility  of  raising. 

The  following  is  the  strain  in  panel-rods  according  to  Fig.  8b  : 


Sect.  =  1,52,  or  1  rod  =  1  X  U". 

The  influence  of  temperature  upon  the  single  systems  of  main 
suspenders  (their  length  being  different)  is  regulated  by  a  link 
connection. 

For  the  compressive  strain  in  the  top  chord  the  rule  for  a  girder, 
sustained  at  both  ends  and  charged  with  an  equally-distributed 
load,  may  be  applied  (see  at  the  close  of  this  chapter),  then, 


SUSPENSION   TRUSS   BRIDGE.  25 

?ooooxj_><£  =  168000  * 

10 

The  compression  in  the  top  chord  is  the  same  all  over. 
For  the  result  we  have  as  momentum  one-half   of  the  entire 
weight  on  posts  at  one-fourth  the  length  of  truss  as  leverage, 

Q      L 

or  Mom.  =        X       ' 

2         4 

which,  when  divided  by  the  depth  of  truss,  gives  the  compression  = 
168000  Ibs.  as  before. 

For  a  single  load,  P,  tit  the  centre  would  be 

P        L 

Mom.  —        X    iy  > 

which,  when  divided  by  the  depth  of  truss,  gives  for  the  compres- 
sion twice  as  much,  or  336000  Ibs. ;  but  for  an  addition  of  the  re- 
sults of  each  single  truss  with  its  single  load,  according  to  x  and 
#!  in  Fig.  9,  it  would  be 

P        '}  L 
Mom.  =        x 

2  8 

this  being  one-half  of  the  result  for  a  single  load,  P,  at  the  centre, 
added  to  one-half  of  the  result  for  an  equally-distributed  load,  Q. 

18.]    The  Fink  truss  (Fig.  18)  is  different  in  principle. 

Whilst  in  the  Bollmann  system  there  are  as  many  independent 
trusses  as  there  are  posts,  in  the  Fink  all  the  trusses  are  dependent 
on  each  other  and  transfer  the  load  toward  the  centre. 

The  centre  post  (king-post)  has  to  sustain  the  compression  of 
one-half  of  the  entire  load  on  the  truss,  including  one-fourth  of  the 
weight  of  the  rib,  the  main  suspenders  (tie-rods)  depending  again, 
as  before,  on  the  depth  of  the  truss. 

*  For  a  simple  compressive  strain  an  area  of  section  of  stretcher  =  9  square 
inches  would  be  sufficient  (18000  Ibs. — safe  load  fur  cast  iron) — the  actual 
dimensions  to  be  taken  by  Hodgkinson's  formula  ou  the  strength  of  hollow 
cast-iron  pillars: 

JJ3.6  J3,6 

W  =  breaking  weight  in  tons  =  44,3  — ; 

therefore,  when  for  a  pillar,  (he  external  diameter,   7),  in  inches,  and  the 
length,  /,  in  feet,  are  known  ;  and  for  six  times  security,  with  the  weight,  W, 
multiplied  by  6,  we  can  define  the  internal  diameter,  d,  consequently  the 
thickness  of  metal. 
3 


26  THE   THEORY    OF   STRAINS. 

The  calculation  of  an  example  in  its  simplicity  will  give  the 
best  explanation. 

Taking  the  same  dimensions  and  the  same  load  as  in  the  calcu- 
lation for  the  preceding  (Fig.  17),  according  to  Fig.  8b  we  have, 

Strain  inA.N or  IN  =  ~  X 

2 

80000  ^,  49 
i.e.,  • X  —  —  196000  IDS., 

the  section  of  wlHch  for  a  value  of  iron  —  10000  Ibs.  per  square 
inch  =  19,6  square  inches.  Thus,  when  two  rods  are  applied,  the 
size  of  each  =  2  X  5  inches. 

Strain  in  kc,  Ak  or  cm  =?  —    -  X  —  =  26000  Ibs. 

2  5 

Section  of  a  single  rod  =  1  X  2$  in.  (full). 
Strain  in  Alor  IE=  —   -  -  ~  =  52000  Ibs. 

Zi         JLv/ 

Section  of  a  single  rod  =  2  X  2J  in.,  or  1  X  5  in. 

For  a  single  locomotive  (weight  84000  Ibs.),  resting  at  cd  on  a 
wheel-base  of  12  feet,  the  vertical  force  for  one  post  at  c  or  d  = 
42000 


2 


=  21000  Ibs.  =  W\ 


then,        Strain  in  tie-rods  —  --  X  —  =  27300  Ibs.  ; 

2  5 

so  the  size  of  rod  kc-  Ak  or  cm  should  be  corrected  to  1  X  2J  in. 

For  the  compressive  strain  on  the  top  chord  (stretcher),  accord- 
ing to  Fig.  14, 

„.       W  AE 
±±  =  —  .  —  , 

2   EN 

and  for  a  vertical  strain  W  =  80000  Ibs.  in  the  centre  post,  us 
before  mentioned, 


2i  LO 

In  this  truss,  as  in  the  Bollmann,  the  compression  in  the  top 
chord  is  the  same  all  over. 

For  this  truss,  when  applied  for  an  over-grade  railroad-bridge,  a 
safe  longitudinal  connection  (bracing  or  tying)  will  be  essential  on 
account  of  the  variable  load. 


?<.,     ovvfit.  *ff 

TT~R 
i 


SUSPENSION    BRIDGE.  27 

To  the  same  category  of  bridges  belongs  the 

SUSPENSION  BRIDGE, 

though,  in  regard  to  mathematics,  very  different. 

The  curve  formed  by  a  chain  or  cable  lies  between  the  parabola 
and  the  catenary,  and  is  very  nearly  an  ellipse.  The  curve  in  a 
loaded  state  approaches  the  parabola ;  in  an  unloaded  state,  the 
catenary.  (Weisbach,  vol.  II.) 

In  the  following  example  the  curve  may  be  considered  a  parabola 
or  the  bridge  in  its  loaded  state. 

The  thesis  is — 

The  vertical  force  at  every  point  of  the  chain  equals  the  weight 
on  the  chain  from  the  point  in  consideration  unto  the  vertex. 
Plate  4,1        So,  when  y  in  Fig.  F  =  25  feet  and  the  length  of 
Fig.  F.J     bridge  =  150  feet, 

Width  =  4  feet ; 

load,  50  Ibs.  per  square  foot,  or  200  Ibs.  per  lineal  foot ; 

maximum  load  =  200  X  150  —  30000  Ibs. ; 

the  vertical  force  at  D  —  200  X  25  =  5000  Ibs. 

Further,  the  horizontal  force  at  every  point  of  the  chain  is  equal, 
and  therefore  equal  to  the  horizontal  strain  in  the  vertex. 

Thus,  when  by  p  is  represented  the  weight  pro  unit  of  horizontal 
projection  =  200  Ibs.,  for  our  example, 

pi2 
I.    //  =  ( — '  which  is  at  the  same  time  the  horizontal  force  in 

~  f  I 

A  and  C,  to  overturn  the  towers,  and  amounting  here  to 
200  X  75' 


20 


=  56250  Ibs. 


II.   8  =  j~  Y#  +  4  A»  -  ^~  X  77,6  =  58500  Ibs. 

III.  *  =  *-.*, 

k 

The  length,  L,  of  chain  results  by  the  formula, 
L  =  2  I  +  i  X  f  =  151,75.* 

o          t 
*  For  more  specifications  I  ought  to  refer  to  Weisbach  and  other  authors. 


28  THE   THEORY    OF    STRAINS. 

For  a  trussed  system  with  two  posts  (queen-posts),  between  sup- 
ports 36  feet,   and  an   assumed   load  =  30000  Ibs.,  equally  dis- 
p,       f.  -j    tributed,  the  distribution  of  forces  <n    tho   bearings   is 
Fig  19J    Denoted  in  Fig.  19,  and  for  the  calculation  of  strain,  when 
compared  with  Fig.  10,  we  fend  for 

r=  w.f-, 

bd 

IQ  04 

T=  10000  X  -      -  =  23700  (approx.) ; 
5,69 

and  for  1",  =  W.  — ;  ^ 

bd 

r,  =  10000  x-  -  =  2ur>7. 

5,59 

The  compression  on  the  vertical  posts  —  10000  Ibs. ;  the  vortical 
pressure  on  supports  —  15000  Ibs. ;  find,  reduced  by  lliose  5000  Ibs. 
directly  sustained  (comp.  Fig.  16),  the  reactive  force  of  supports, 
signified  by  D  =  10000  Ibs. 

Upon  this  structure,  the  Metfiod  of  Moments  being  applied  (see 
Preface),  we  suppose  a  section  separated  from  the  original  by  a 
cut,  */. 

Considering  the  forces  acting  upon  such  a  section,  we  form  the 
equation  of  equilibrium  for  a  suitable  point  of  rotation,  by  solution 
of  which  we  find  the  strain  in  the  member  in  question  ;  and  observe 
the  rule,  that,  for  a  strain,  Y,  the  point  of  rotation  ought  to  be  chosen, 
in  the  intersection  of  x  and  z,  making  their  lever  —  0,  when  these  are. 
the  members  of  the  structure,  separated  by  a  cut,  «•</,  likewise  a«  Y. 
9ft  ,  But  when  by  st  only  one  member,  excepting  Y,  is  separated, 
we  lay  tho  point  of  rotation  on  the  next  joint,  as,  per  ex- 
ample, for  Fin  b,  Fig.  20. 

F.  5,07  =  10000  X  12, 
or,  I.,      0=--  -Y.  5,07  +  10000  X  12  (rotation  round  b) ; 

r=L«ow>x_i2  =  + 23700,    '    ';• 

and  because  in  the  following  the  form  of  equation  always  will  be 
kept  similar  to  I.,  the  forces  in  their  aim  to  turn  to  the  left,  like 
Y  round  b,  will  be  signified  by  — ,  the  same  as  a  compressive 
stniin  ;  and  the  forces  to  the  right,  like  the  hands  on  a  watch,  or  D 
round  b,  will  be  Dignified  by  -(-,  the  same  as  a  tensile  strain. 


SUSPENSION    BRIDGE.  29 


21.]     According  to  this  we  have  for  Z,  Fig.  21, 

Q  =  Z.  5,59  +  D  .  12  (rot.  r  .  d\ 
MOOOXI2 


5,59 

and  for  F,,  Fig.  22, 
22.]  0  =  —  Fj.5,59  +  Z>.12(rot.  r.6); 

Fi=  10000X12 

5,59 

When  a  diagonal,  s,  sustains  the  parallelogram,  so  that  by  a  cut, 
$1,  three  members,  Z1?  Yl  and  s  are  separated,  we  have  for  the  defi- 
nition of  s  the  point  of  rotation,  as  before  mentioned,  in  the  inter- 
section of  Zl  and  Yl  ;  but  Zv  and  F,  in  their  direction  are  parallel, 
and  therefore  without  intersection  at  all.  In  this  case  (the  same  as 
for  diagonals  in  girders  with  parallel  top  and  bottom  flanges)  we 
OQ  -]  suppose  a  point  of  rotation,  0,  at  any  distance  in  the  direc- 
tion (axis)  x,  and  find  thus  from  Fig.  23,  where  x  =  oo  ,  or  in- 
finite —  i.e.,  the  lever  of  all  forces,  acting  in  a  vertical  direction 
upon  the  section  =  cc  . 

0  =  —  *  .  oo  sin  <p  —  D  .  oc  -f  p  .  oo    (rot.  r  .  0)  ; 
or,  because  oo  is  a  factor  of  each  part, 

0  —  —  s  .  sin  y  —  10000  +  10000. 

In   this  equation,  s  .  sin  .  <?  (comp.  Fig.  56  on  semi-girders)  is 
9,  1    the  vertical  component  of  a  (Fig.  24),  and  acts  right-angled 

to  the  axis,  x,  like  D  and  p. 
The  angle  <f  for  our  example  —  25°  ; 

sin  <p  =  0,422  (see  table), 

und  therefore    0  =  —  s  X  0,422  —  10000  +  10000, 
or  s  =  0, 

showing  that,  as  in  Fig.  19,  the  diagonal,  s,  is  without  any  strain, 
and  only  useful  for  preventing  dislocation.     (Compare  parabolic 
25  1    S"'c^ers'  to  which  this  case  is  similar,  because  a  parabola  can 
be  constructed  through  the  sustaining  points  a,  d,  e  and  c,  and 
therefore  differs  in  this  from  Figs.  25  and  (53.) 
n/>  -i        For  a  reversed   structure  (Fig.  2(i)  the  strains  will  be  the 
same,  but  of  reversed  signs. 

In  the  following  calculations  of  rooTs  and  bridges  it  will  be  shown 


30  THE  THEOEY   OF  STRAINS. 

that  the  Method  of  Moments  is  thoroughly  applicable,  leading  di- 
rectly and  in  the  most  comprehensive  manner  to  distinct  results ; 
but  for  a  preliminary  estimate  of  strain  in  the  top  and  bottom 
chords  at  the  centre  of  the  structure  the  most  simple  way  to  define 
this  strain  may  be  stated  by  the  following: 

As  the  flanged  girder  in  Fig.  27,  charged  with  an  equally-dis- 
tributed load,  O,  will  be  exposed  at  its  centre  to  the  same 
27  1 
on'l    exertion  as  the  girder  in  Fig.  28,  fixed  at  the  centre,  the 

moments  will  be  for  both,  when  V=  depth  of  girder  and 
I  =  length  (the  load  being  equally  distributed). 

Mom  =  --  •  -  =  V.  x  (rot.  r .  o) ; 

~         TC 

£  ' 

2    4       IQ.l 

— — —       — - —    =  compression  or  tension  in  flanges ; 

so  Q  =  30000  Ibs. ;   V  ==  5,59';  and  I  =  36'.     - 


and  -  —  24150  Ibs.,  the  strain  in  the  chords. 

5,o9 

It  is  to  be  observed  that  the  result  is  rather  too  high  wtien  ap- 
plied upon  a  truss  with  few  panels,  as  in  Fig.  26,  on  account  of  the 
reactive  pressure  of  the  support,  diminished  by  the  partition  of  the 
direct  load  on  this  place. 

On  account  of  its  being  a  very  convenient  method  we  recommend 
it;  and  in  the  following  bridge  skeletons  we  refer  to  it  very  fre- 
quently. (See  D,  Example  I.,  and  Sect  II.,  note.) 
29.]  ^or  a  8ir<3er  Charged  with  a  single  weight  at  the  centre 
(Fig.  29),  we  make  a  comparison  with  Fig.  30,  and  find  for 
both 


.1  p  i 
30.]     and  -—f —  =  horizontal  strain  ; 


t.  e.,  compression  or  tension  in  the  top  or  bottom  chords,  the  for- 
mation of  moments  for  a  point  of  rotation,  o,  being  very  compre- 
hensive. 


When  P=  15000  Ibs., 


e>.. 


B.    ROOF  CONSTRUCTION.  31 


but  V  =  5,59,  and  I  =  36  (as  before), 


so  the  Mom  =  *  f  =  135000> 

15000       36 

22 

and  -  =  24150  Ibs.,  the  strain  in  flanges  ; 

5,59 

showing  that,  for  the  same  exertion,  a  beam  loaded  with  a  single 
weight,  P,  at  the  centre  can  bear  only  one-half  of  an  equally- 
distributed  load,  Q. 

[Plates  3,  4  and  5—  embracing  Figs.  13  to  30.] 


B.    EOOF    CONSTRUCTION. 

For  small  and  not  complicated  roofs,  experience  is  a  common  and 
in  general,  also,  a  sufficient  guide.  But  experience  is  very  limited, 
and  not  every  constructor  has  opportunity  and  time  to  acquire  it. 

The  true  and  acceptable  guide  for  a  safe  practice  will  always  be 
the  calculation  ;  and  since,  especially  for  complicated  and  more 
extensive  combinations,  the  application  of  mechanical  science  luis 
become  unavoidable,  the  following  compendium,  leading  from  the 
simplest  to  the  most  complicated  structures,  will  give  for  almost 
every  purpose  sufficient  information. 

By  construction,  when  EG  represents  the  weight  at  the  centre 
of  gravity,  G  (Fig.  31"),  the  body  will  be  at  rest  when 

Fi^Sl*  J    the  plane  DF  is  rigllt-{lngled  to  tlie  line  DE'i  GE  being 
horizontal  —  i.  e.,  right-angled  to  the  vertical  line  EG. 

Let  EG  be  an  assumed  length,  then  in  the  parallelogram  of 
forces  the  intensity  of  EK  and  EH  is  measured  in  proportion  by 
the  same  rule. 

For  an  angle,  13  =  25°  of  a  rafter  with  the  horizontal  line 
op  -i  (Fig.  31b  —  similar  to  Fig.  16  reversed)  leaning  with  the  top 
end  against  a  wall,  the  heel  at  A  being  morticed  ;  we  then 
have 

-  =  0,422; 


sin2  13  =  0,17; 
sin  2/3  =  bin  50°  =0,766; 


32  THE   THEORY   OF   STRAINS. 

^•  =  cos  t  =  -i|-  =  0,906; 
ac  13,24 

cos2 ;?  =  0,82. 
When  W  is  an  equally-distributed  load  of  20000  Ibs.,  then 

*=  ffl  =  ]rxd     goooo  x«  =21467  lbs        . 

2        A           2           o,o9 

The  vertical  force,  F,  at  the  top  of  the  rafter  =  0,  and  the  ver- 
tical  pressure  F2  of  the  heel  =  20000  lbs.,  or  equal  to  the  entire 
load. 

Also  the  vertical  force,  FI  at  the  centre  =  20000  lbs. 

The  pressure  in  the  rafter  itself  (compression)  = 

W        1  1  3  24 

-  X  -  =  10000  X  -  —  =  23685  (23700  lbs.). 
2        h  5,59 

The  entire  pressure,  ^R,  df  rafter  toward  the  support, 

R  =  Vvf+H?  =  29300  lbs. 

Its  direction  can  be  constructed  in  making  K—  2h  —  11,18  feet, 
HI  and  F2  forming  the  sides  of  the  parallelogram. 

When  P  —  10000  lbs.,  a  single  weight  at  centre  of  rafter, 

TT  TT    •*         **   -I  n7°Q  . 

Jtl  =  Xli  —     —   .  —     _  lyjidfj 

2    h 
K«=o, 

and  V,  —  F,  =  10000  lbs. 

When  in  Fig.  31",  by  FM  the  weight  of  the  body  AB CD  is  rep- 
o-jo  -i  resented,  then  FN,  the  force  toward  the  wall,  results  in  the 
horizontal  and  vertical  forces  CH  and  (7F. 

FL,  the  force  acting  perpendicular  to  the  plane  SI,  in  the  direc- 
tion BF. 

G,  the  centre  of  gravity. 

GF,  vertical. 

GFl  CB,  or  <  FCB  =  90°. 

BI  l  FB,  then  BI  is  the  direction  of  the  plane  required  to 
make  the  body  at  rest. 

Also  FN  =  P  =  *-Q  CM  *• 

H  —  P  sin  oc  =  i  G  sin  2  «. 


M=JVilZs*<Z^ 


\ 


B.     ROOF    CONSTRUCTION.  33 

oid  -,        Let  the  sloping  body  ABCD  (Fig.  31d)  be  supported  by  a 
wall  at  its  lower  end,  D,  which  coincides  with  the  surface  of 
the  body  ; 

Let  G  be  again  the  centre  of  gravity  ; 

It  is  required  to  cut  a  notch  out  of  the  body  at  the  upper  end, 
C,  so  that  it  may  rest  upon  the  top  of  a  wall  which  is  made  to  fit 
the  notch. 

Make  GE  vertical  ; 

From  D  drzwDE  l  to  CD  ; 

Join  EC,  and  draw  CF  at  right  angles  to  it  ;  then  the  notch  ufc 
C  being  cut,  the  body  ABCD  will  be  at  rest. 
q-,e-i        A  body,  ABCD,  resting  on  supports  (Fig.  31"),  will  only 

DE  EL 

produce  the  vertical  strains  —  —  •  Q  and  —  —  •  •  Q  at  the  sup- 

ports. DL  DL 

Plate  7,1      For  Figs.  32,  33  and  34  we  have,  again, 
Fig.  32.J 

be  .    ab  ,    be  ab 

—  =  sm  p  ;  —  =  cos  p  ;  —  =  tang  p  ;  —  ==  cotg  p  • 

ac  ac  ab  be 

sin  2  p  =  sin  50°  =  0,766; 

and  when  upon  each  rafter  W  =  20000  Ibs.,  equally  distributed,  we 
have  for  Fig.  32, 

W  rl        W 
H=  H1  =  —-j-  =  —  •  cotg  /9  =  21467  Ibs. 

2     A          — 

W  I 

The  compression  in  the  rafter  =  —••-==  23685  Ibs., 

2     fi 


and,  again,  R  =  T/JJ^  _j_  y*  =  29300  Ibs. 

For  an  angle  /9  =  26°   34';  the  horizontal  thrust  will  —  20000 
Ibs.  —  i.  e.,  the  same  as  the  entire  weight.     (Comp.  Fig.  16.) 
go  -i        For  Fig.  33  is,  as  the  rafter  in  a  vertical  direction,  sus- 
tained on  the  top, 

JBT=  J3i  =  ^-sin  2-/J  =  5000  X  0,766  X  3830; 

4 

V=^-  cos3  0  =  10000  X  0,82  —  8200  ; 

Z 

W 
V,=  Z-  (1  +  8in'/3)  =  10000  (1  +  0,17)  +  11700; 


34  THE  THEORY   OF  STRAINS, 

and  the  compression  in  the  rafter, 

—  •  sin  0  =  10000  X  0,422  =  4220  Ibs. 
2 

For  Fig.  34,  when  the  rafter  is  sustained  at  the  top  by  a  vertical 
post, 

H  =  H,  =  —  •  sin  2  /3  =  3830 ; 
4 

34.]  V  =  W  cos2  /3  ==  20000  X  0,82  =  16400 ; 

W 

y^lL(lJr  sin*  ft  —  10000  (1  -f  0,17)  =  11700 ; 

z 
and  the  compression  in  the  rafter  = 

W 

—  -  sin  /9  =.-  4220  Ibs. 

In  the  cases  in  Figs.  33  aud  34  the  post  relieves  the  tie-rod  or 
(as  here,  in  the  absence  of  a  tie)  the  wall  from  a  part  of  the  thrust 
of  the  rafters,  and  compared  with  the  truss  in  the  preceding  we 
see  that  the  king-post  acts  in  a  different  way. 

The  different  combinations  of  this  single  hanging-and-thrust 
construction  may  be  disregarded,  and  by  the  following  calculations 
the  method  of  moments  thoroughly  applied. 

A  rafter  being  constructed  in  this  way  (Fig.  35),  without  con- 
nection between  the  point  e  and  the  horizontal  tie-rod  aa,  there  are 
two  different  systems,  that  of  a  trussed  beam,  aec,  similar  to  Fig.  16, 
or  -|  and  that  of  a  single  triangle  like  Fig.  32,|and  it  is  as  there  for 
the  same  angle,  /?,  and  a  load  =  20000  Ibs.  upon  each  rafter ; 
the  horizontal  thrust,  H,  at  the  top,  as  also  the  tension  in  the  hori- 
zontal tie-rod  =  21467  Ibs.  The  trussed  rafter  is  understood  to  be 
calculated  like  Fig.  16  in  combination  with  Fig.  31b. 

When  the  weight  of  structure,  snow  and  wind-pressure  upon  a 
gg  -,  roof  (Fig.  36),  for  one  square  foot  of  horizontal  projection, 
in  all  =  50  Ibs.,  and  the  width  between  the  supports  or  AC  — 
40  feet,  the  distance  of  rafters  =  10  feet,  then  20000  Ibs.  is  the 
entire  load,  or  10000  Ibs.  upon  each  rafter,  supported  at  A,  B,  C, 
D  and  E. 

The  pressure  of  one-half  of  the  weight,  or  10000  Ibs.,  on  the 
supports  is  counteracted  by  the  direct  load  of  2500  Ibs.  It  is  there- 
fore the  reacting  force,  D,  only  7500  Ibs.  (See  Howe  Truss, 
Sect.  I.,  D.) 

For  a  section  separated  by  st  we  can  define  at  once  the  strains  in 


B.     ROOF    CONSTRUCTION.  35 

z,  y  and  x1}  so  for  xlf  considering  the  forces  acting  upon  this  section 
and  the  point  of  rotation  in  the  intersection  of  y  and  z  or  F. 
(Fig.  36a.) 
36".]  0  =  x,  .  a  +  D  .  20  —  5000  X  10, 

(Comp.  Fig.  20,  Equat.  I.) 

The  arm  or  lever,  a,  can  be  measured  near  enough  from  a  skele- 
ton =:  9,6'. 

Besides,  for  the  calculation  of  a  we  have  for  the  angle  ABG 
(Fig.  36), 

AG=  GB  .  tang  <  ABG, 


t.  e.,  <ABG  =  53°  7',  and  sin  53°  T=  0,8  ; 

a  =  BF  sin  <  ABG  =  12  X  0,8  =  9,6  ; 

therefore      0  =  x,  .  9,6  -f-  7500  X  20  —  5000  X  10  ; 

100000 
xl  =  --    -  =  —  10412  11)3. 

9,6 
For  z  (rot.  r  .  E,  Fig.  36a), 

0  =  —  z  .  6  +  7500  X  10, 
*  =  +  7-f»  =  +  12500; 

and  for  y  (rot.  r  .  A,  Fig.  36a), 

Q  =  y.c+  5000.10; 
Q  —  y.  10,75  +  50000, 
50000 

y  =  -7o^  =  -4650- 

Plate  8,  1     For  the  strain  in  x  (rot.  r  .  F,  Fig.  36b)  is, 
Fig.  36b.J  0  =  x    9)6  +  750Q  x  20> 


or  x  =  —  -         -  =  —  15625, 

J,o 

In  regard  to  the  vertical  V,  we  use  for  its  definition  the  strain  of 
the  joining  brace,  ^  —  —  10412,  and  make  st  a  curved  line  ;  then 
we  have,  for  a  rot.  r  .  D  (Fig.  36b), 

0  =  —V.  10  —  (—  10412)  .  10,9, 
or  0  =  —  F.  10  +  113490; 


36  THE   THEORY    OF    STRAINS. 

_    •       113490 
t.«.,  V=  -\  --  —  —  =  +  11349. 

37.]     The  results  are  combined  in  Fig.  37. 

For  Fig.  38,  the  entire  load  (equally  distributed)  again  being 
20000  Ibs.,  the  depth  15',  and  between  the  supports  40'. 
When  here  the  cut  st  separates  the  line  %$&,  we  have  for  #j 
(rot.  in  the  intersection  of  t/jZi,  or  F,  Fig.  39), 

0  =  Xl.  9,1  —  5000  X  5*  +  7500  X  15*  ; 


39.]  Il=_  9752. 

y,i 

For  yi  (rot.  r  .  A\    0  =  —  yl  .  15  -f  5000  X  10, 

y,=  +  522p=  +3,333; 

and  for  z,  (rot.  r  .  C), 

0  =  -^.  15  —  5000  X  10-^7500  X  20; 

2,  =  -f-  6,666. 

40.]     For  a  section,  st,  through  x  and  z,  we  have  for  x  (Fig.  40), 
0  =  x.  9,1  +  7500  X  151  (rot.  r.F}; 
116250 

TOT     "12774; 

and  for  z  (rot.  r  .  D\ 

0  =  —  2  .  7,4  -f  7500  X  10  ; 


8 

For  y  (rot.  r  .  A)  is,  0  =  y  .  12,5  +  5000  X  10  ; 

,  =  -^  =  -4000.  ' 

41.]     The  results  combined  in  Fig.  41. 

42.]        When  the  figure  before  is  changed  in  the  depth,  like  Fig. 

42,  we  have  the  following  equation  : 
0  =  Xl  .  5,8  —  5000  X  3,25  +  7500  X  13,25  (rot.  r  .  F,  Fig.  43)  • 


43.]        For  tlic  definition  of  y\,  the  intersection  of  a*  and  \  will  be 
in  G,  and  it  is  for  G  as  rotation, 

0  =  —  y,  .  8,25  4-  5000  X  6  -f  7500  X  4  ;      (Fig.  43.) 


-l^rT'i 


B.     ROOF   CONSTRUCTION.  37 

H  yi=+?2005 

8,25 

For  Zi  (rot.  r  .  (7)  we  have 

0  =  —  z,  .  12  —  5000  X  10  +  7500  X  20; 

ooooo 


J.  ^J 

44.]     For  a  section,  s£,  through  #  and  z  (Fig.  44),  it  i 
0  =  x  .  5,8  +  7500  X  13,25  (rot.  r  .  F)  ; 
993750 


5,8 
and  0  =  —  z .  5,2  +  7500  X  10  (rot.  r .  D) ; 

75000 

~5j~  '°; 

and  for  y,        0  =  y  .  12,5  +  5000  X  10  (rot.  r.A); 

50000 

y  = —  —  4000. 

12,5 

45.]     The  results  combined  in  Fig.  45. 

Plate  9,1       For  the  definition  of  X  in  Fig.  46,  the  point  of  rotation 

Fig.  46. J   in  E,  or  in  the  intersection  of  Y  and  Z,  will  be  from 

Fig  47. 

47.]  0  =  X.  x—  P.CE+  D.AE; 

P.CE—D.AE 

or  X= 

x 

For  Y  we  choose  A,  or  the  intersection  of  X  and  Z,  as  the  point 
of  rotation,  and  the  equation  will  be 

0  =  —  Y.  y  +  P.  AC  +  Q .  AE, 
or  Y=  — 


y 

and  in  the  same  way  for  Z,  rot.  r  .  H. 


z  -^  — 


It  will  tiot  be  necessary  to  show,  by  repetition  of  the  foregoing, 
the  equations  i'ur  the  other  parts  of  the  structure. 


38  THE  THEORY   OF  STRAINS. 

4S  -,        In  more  complicated  systems  (Fig.  48),  it  may  happen  that 
by  a  cut,  st  (which  can  be  made  curved  as  \vell  as  straight), 
different  braces  or  rods  are  spared,  like  FG,  D  G  and  DE. 

In  this  case  it  is  possible  to  come  to   a  direct  result  when  st 
only  can  be  laid  so  that  all  the  braces  or  rods  cut  by  st  meet  at 
one  point,  except  that  one  whose  strain  is  in  question. 
49.]     So  for  the  strain  V  in  FG  (rot.  r  .  H,  Fig.  49), 

—  E.r; 


V- 


FH 

50.]     In  the  same  manner  the  strain  U  in  DG  (rot.  r  .  H,  Fig.  50), 
Q=  U.u  —  E.r; 


u 

thus  we  find  also  the  strain  in  KT  and  LT.        . 
r-i  -i        Being  by  the  foregoing  in  possession  of  a  value  for  U  in 
1    DG,  we  find  for  the  strain  X  in  DF,  Y  in  DE,  and  Z  in  CE 
the  following  equations  from  Fig.  51  : 

0  =  X.DE  +  U.v  —  Q.  NO  —  P.  MO  +  W.  AO  (rot.  r  .  E}  ; 
0  =  Y.  AD  +  U.  I  +  Q  .  AN+  P.  AM  (rot.  r  .  A)  ; 
0  =  —  Z  .  z  +  W.  AN—  P.  MN  (rot.  r  .  D)  ; 

each  6ne  enabling  us  to  obtain  a  direct  result  for  the  strain  in 

question. 

Plate  10,1        For  a  roof  (Fig.  52),  the  weight  of  which,  11,3  Ibs. 

Fig.    52.  J     per  square  foot  of  its  horizontal  plan,  may  be  calcu- 

lated 20  Ibs.  for  wind  pressure  and  snow,  making  together  31,3  Ibs. 

per  square  foot. 

The  distance  of  rafters  being  15J  feet,  the  width,  100  feet,  makes 
for  each  rafter  15  J  X  100  X  31,3  =  48000  Ibs.  (approx.). 

The  load  at  each  apex,  therefore,  will  be  —     —  =  6000  Ibs.,  the 

8 

distribution  of  which  is  shown  by  the  skeleton. 
For  the  reactive  force  on  the  supports  is  again 

D  =  24000  —  3000,        or  D  =  21000  Ibs. 

There  are,  in  all,  seven  times  6000  Ibs.  acting  downward,  and 
twice  21000  Ibs.  acting  vertically  upward  upon  the  system. 


B.     ROOF   CONSTRUCTION.  39 

go  -j        The  section,  A,  s,  t  (Fig.  53),  kept  in  equilibrium  by  the 
replaced  forces,  x,  y  and  z,  may  be  regarded  first  as  a  lever 
with  the  fulcrum  at  D ;  then  the  strain  in  x  for  the  middle  sec- 
tion is 
0=  x .  18,6  +  21000  X  50  —  6000  X  12,5  —  6000  X  25  —  6000  X 

37,5, 
or  x  =  32300  Ibs. ; 

and  in  y,  when  A  is  the  point  of  rotation, 

0  =  y .  38,4  +  6000  X  12,5  +  6000  X  25  +  6000  X  37,5  (rot.  A)  ; 

y  =  — 1 1700  Ibs., 
and 

0^  —  2.15+21000x37,5  —  6000X12,5  —  6000X25  (rot. r.E.), 

Z  =  +  37500  Ibs. 
54.]     For  V  in  Fig.  54  the  rotation  also  round  A  is 

0  =  —  V.  37,5  +  6000  X  12,5  a±  6000  X  25 ; 

V=  +  6000  Ibs. 
For  the  other  members  in  Fig.  52, 

0  ==  a?, .  13,9  +  21000  X  37,5  —  6000  X  12,5  —  6000  X  25 
(rot.  r.F*); 

xl=  —  40400 ; 

0  =  yl .  23,5  +  6000  X  12,5  +  6000  X  25  (rot.  r.A); 

yi=— 9570; 
0  =  —  z, .  10  +  21000  X  25  —  6000  X  12,5  (rot.  r.G); 

Zl  =  +  45000 ; 
0  =  —  Fi .  25  +  6000  X  12,5  (rot.  r.A); 

F,  =  +  3000 ; 
0  =  x2 .  9,3  +  21000  X  25  —  6000  X  12,5  (rot.  r .  H )  ; 

x2  =  —  48400 ; 
0  =  y, .  9,3  -f  6000  X  12,5  (rot.  r.A)', 

y*  -=-8100; 

0  =  —  z, .  5  +  21000  X  12,5  (rot.  r .  /) ; 
z,  =  +  52500. 

For  the  strain  in  x3  we  choose  a  convenient  point  for  rotation  in 
the  line  z,  per  Example  D,  Fig.  55. 


40  THE   THEORY   OF   STRAINS. 

55.]     The  equation  in  this  case  will  be 

Q  =  x3.18,Q  +  21000  X  50; 

jT8  —  —  56500. 

The  only  strain  not  directly  deducible  is  U  in  the  vertical  line 
CD  at  the  centre. 
As  in  Fig.  36,  we  use  the  strain  of  the  joining  brace, 

x  =  —  32300  Ibs. 

56.]     For  B  as  the  point  of  rotation  (Fig.  56),  our  equation  is 
0  =  —  U.  50  —  6000  X  50  —  (—  32300)  .  37,2  ; 

U=  18000  Ibs. 
67.]     The  results  combined  in  Fig.  57. 

The  weight  and  load  of  a  roof  (Fig.  58)  being  estimated,  in- 

f-ft  -.  k  eluding  wind-pressure  and  snow,  to  50  Ibs.  per  square  foot 

of  its  horizontal  plan,  the  distance  of  rafters  being  12  feet, 

and  the  space  between  the  walls  99  feet,  which  gives  50  .  12  .  99  — 

59400  Ibs.  for  one  rafter,  or,  in  round  figures,  60000  Ibs. 

The  calculation  of  the  top  structure  can  be  made  as  in  the  pre- 
ceding example  (Fig.  36). 

In  the  main  construction  are  six  supporting  points,  charged  as 
P-o-i    in  Fig.  58.     The  top  structure  transmits  one-third  of  the 

entire  load,  or  on  each  post  10000  Ibs.  to  the  apexes,  ff. 
Each  wall  has  to  bear  30000  Ibs.  ;  and  after  subtraction  of  the 
direct  load  the  reactive  force  is  26666  Ibs.,  or,  by  calculation, 

_  6666  (11  +  22)       13333  (33  +  66)       6666  (77  +  88)  t 
99  99  99  ; 

D  =  26666  Ibs. 
59.]     For  the  strain  x3  we  have  in  Fig.  59, 

0  =  —  *,  .  21  —  13333  X  161  —  6666  (27*  +  381)  -f-  D.  491 

(rot.  r  .  /i)  ; 
or,  also, 

0  =  —  x,  .  21  —  6666  (11  +  22)  +  26666  X  33  (rot.  r  ./)  ; 
659967 


Further, 

0  =  Zt.  21  —  13333  X  161  —  6666  (271  +  381)  +  D  .  491 
(rot.  r  .  g). 


%.jr      *££•/£  I  %*  .ir| 

.^-^Q^  V;/': 


B.     ROOF    CONSTRUCTION, 

659967 


and    0=y3.  39,5  -f  13333  X  33  -f  6666  (11  +  22)  (rot.  r  .  o)  ; 
y?  =  _  16708. 

The  tie-rod,  gh,  transmits  the  strain  to  the  top  flange,  and  is  herg 
sustained  by  the  counter-brace,  eh. 
60.]     From  Fig.  60  is 

0  ==  —  x3  .  14  —  6666  X  11  +  D  .  22  (rot.  r  ,  ef)  ; 

513304 
xt  =  -  =  ooboo  ; 

0  =  z3  .  171  —  6666  (11  +  22)  -f  26666  X  33  (rot.  r  .  «)  j 

II     —  s—  «••  •  ;:";:;: 

0  =  2/2.  26,7  +  6666  X  11  +  6666  X  22  (rot.  r  .  o)  ; 

7/2  =  —  8223. 
61.]     Fig,  61  gives  the  equations, 

0  =  —  Xl  ,  7  +  26666  X  11  (rot  r  ,  6)  ; 

^  =  +  41902  ; 

0  =  22,13  —  6666  X  11  +  26666  X  22  (rot.  r.c); 
22  =  —  39485  ; 

0  =  y,  ,  13  +  6666  X  11  Crot-  f  ^a)  5 
j/,^  —  5640; 

»nd  for  «i  we  find  from  the  same  figure, 

0  =  .«!  .  13  +  26666  X  22  ; 

zl  =  —  45125. 
62.]     For  the  strain  in  tie-rods  we  find  from  Fig,  62. 

0  =  —  F,  ,  33  +  6666  .  (11  +  22)  (rot.  r  ,  a)  ; 

V3  =  +  6666  ;  (Comp.  Fig.  68.) 

0  =  —  V.  .  22  +  6666  X  11  (rot.  r  .  a)  ; 

F2  =  +  3333  ; 
Q  =  —  Fi.11  +  0  (rot.  r.a); 

Fi  =  0  (and  is  therefore  not  egsential)^ 


42  THE  THEORY  OF  STRAINS. 

^ 

go  -i        The  strain  in  Vt  at  the  centre  rod,  according  to  8b,  can  be 

defined  thus : 

01 
V<  =  2  X  -=?-  X  16708  =.26200  Ibs. 

The  results  are  combined  in  Fig.  63. 

When  in  Fig.  64  the  rafters  are  trussed — i.  e.,  stiffened  by  a 
»A  -i  king-post  at  b — there  will  be  only  four  supporting  points  in 
the  main  construction,  because  the  load  in  this  case  is  trans- 
ferred to  the  wall. 

9999X22       13333(33  +  66).    9999X77      9qqq9 
~99~  ~~99~  ~99~~ 

Plate  12,1      Further  in  Fig.  65, 
Fig.    65.J 

0  ==  xz .  21  —  13333  X  16J  —  9999  X  27£  +  23332  X  49* 
(rot.  r .  Ji)  ; 

x,  —  31427 ; 

0  =  z3 .  21  —  13333  X  16J  —  9999  X  27*  +  23332  X  49* 
(rot.  r .  g) ; 

s3  —  —  31427 ; 

0  =  y, .  39,5  -f  9999  X  22  +  13333  X  33  (rot.  r .  a) ; 

y,  =  — 16708. 
66.]  And  from  Fig.  66, 

0  =  —  *! .  14  +  D.  22  =  —  s, .  14  +  23332  X  22  (rot.  r .  d) ; 

X!  =  +  36665 ; 
0  =  02 . 171  +  23332  X  33  —  9999  X  11  (rot.  r.e}-, 

zt=  -  37180; 
0  =  y, .  26,75  +  9999  X  22  (rot,  r .  a)  ; 

y,=  -8223. 
67.]  For  Z  we  have  from  Fig.  67, 

0  =  ^  .  13  +  23332  X  22  (rot.  r.e); 

Z,=    -39485. 
68.]     See  the  results  in  Fig.  68  combined. 

The  strain  F4  =  26200  Ibs.  can  be  defined  independently  of  the 
Method  of  Moments  by  the  parallelogram  of  forces,  as  in  Fig.  63, 
already  shown, 

F4=2(—  16708).  cos  oc; 


x.,     • 


>'  \      ^SV    9 


SEMI-GIRDERS   LOADED   AT   THE   EXTREMITY.  43 

and  when  by  means  of  counter-braces,  e,  h,  the  top  chord  is  re- 
lieved from  the  strain,  so  that  one-half  to  each  side  is  transported 
to  the  tie-rods,  e,f,  then  here  the  strain  will  increase  to  13000  -f- 
6666  =  19666  Ibs. 

In  a  combination  of  rafters  (Figs.  69,  70),  the  pressure  of 
Y^'  I    the  end  rafters  upon  the  wall  results  in  an  outward  horizontal 

and  vertical  force. 

Different  from  this  is  the  action  of  the   intermediate   rafters, 
being  similar  to  an  oblique  bridge-truss,  sustained  at  the  top  chord. 
The  horizontal  force  at  the  heels  of  the  intermediate  rafters  is 
opposed  to  the  horizontal  force  of  the  end  rafters. 

[Plates  6,  7,  8,  9,  10,  11  and  12— embracing  Figs.  31  to  70.] 


C.    SEMI-GIRDERS. 

I.    SEMI-GIRDERS  LOADED  AT  THE  EXTREMITY. 

pi  ,  i  o  -|        As  the  most  simple  presentation  for  a  weight,  W,  the 
Fijr.    7lJ    stress  in  struts  and  tie- rods  is  inscribed  in  Figs.  71  to 

74,  and  the  parallelogram  of  forces  connected. 
73.]      To  compute  in  Fig.  73  the  stress-  in  the  lower  flange,  we 
have  Z 

de  2 

—  =  sec  QC,  and =  • —  sm  oc, 

df  —  W.  sec  cc 

& 

or  -  =  —  W.  sec  oc.  sin  cc ; 

Z= — W.  sec  a.  since — W.  sec  cc  sin  QC, 
or  Z  =  —  2  W ' .  sec  cc  .  sin  QC  ; 

,•  .  tang  oc 

and  since  sin  cc  — a — , 

sec  a 

Z^  -2TF.seca^^  =-2W.  tang  oc 
sec  oc 

(much  easier  determined  in  Fig.  77  by  the  Method  of  Moments). 

From 'Fig.  73  and  the  following  we  see  that,  for  a  load  at  the 
extremity,  the  diagonals  are  strained  equally  and  alternately  with 
tensile  (-J-)  and  compressive  ( — )  strains.  (Comp.  Fig.  23.) 


44 


THEORY   OF   STRAINS; 


But  the  strain  in  the  flanges  increases  toward  the  support  ift 

each, 

2  W.  tang  «:  , 

where  «:  is  the  angle  of  diagonals  with  a  vertical  line* 
75.]        For  a  better  presentation  of  this,  see  Fig.  7o,  and  for  the 
calculation  apply  the  Method  of  Moments. 

«g  -i        When  by  a  cut,  st,  a  section  of  the  structure  is  separated; 
77*  J    we  nave  f°r  the  flanges  as  equation  of  equilibrium  (Figst 
76  and  77), 


0  =  - —  Xi .  cb  -f-  W.  ca  (rot.  r .  i), 
or  for  cb  —  Av  and  ac  =  I ; 

x=  +  W.l-  =  +  l 
h 

78]      and  by  Figs.  76  and  79 : 
79] 

0  =  —  x2 .  h  +  W.  3  f  (rot.  r .  e)  ; 
a-2  =  -f  3TF-  -  =  3TT.  tang  oc. 

In  the  same  tnanner  is 
,0  =  —  a?, . h  +  W. 5 1  (rot.  r. $0  ; 
,  i 


0  =  +  2, .  h  +  IT".  2  /  (rot.  r .  d)  • 


0  = 


0  = 


4-F;<n(rot.r./)! 

^_ 

A~ 


PI  t   14  1        ^  wrought''ron  crane  (Fig.  A),  constructed  of  braces 
Fig    A.  J   wittl  ^n^-Jomts»  may  be  loaded  at  the  extremity  with 
30000  Ibs.  =  P;  so  is  (for  the  dimensions  noted  in  the 
skeleton)  the  horizontal  strain,  8,  in  a  and  6. 

0= --5.  6  + P.  12  (rot.  r.d); 

fif  =  60000  Ibs ; 

B.]     and  for  the  other  members  we  have 
0  =  —  *i  •  0,75  -f  0  (rot.  in  the  intersection  of  ft  and  4*  or  g,  Fig.  B)  | 

o  =  ft.i,l— P.  2,1  (rot.  r.O; 

,    "0000X2,1 
^  ~   ' T; —  °7272  5 


If 


u 


y^fcjhs*.  ^ 

^\     *^s^ 


if  3t%       |t 


fi^. 


k^T  t  ,^/t. 


-J  "1  ,-'  ./« 

-;/_  J      ^,  --:--• ;."'"' 


%;>k\ 

'•»^.  *v^* 


TV 


•/ 


> 


?: 

X          ^ 


LOADED  AT  EACH  Affix.  45 


=  ^.  1,8  +  P.  4  (rot./); 
30000.4 


1,8 
0  =  —  z,  .  1,99  +  P.  4  (rot.  r  .  e,  Fig.  C)  ; 

120000 
2,  =  -  —  +  oOdOi  ; 

1,99 
0  =  2/2  .  4,4  —  P%  1,2  (rot.  r  .  /O  J 


>.8  (rotr.d); 

:r2  =  —  80000 ; 
ft.]  0  =  —  es .  3,2  +  P.  8  (rot.  r .  c,  Fig.  D)  { 

23  =  75000 ; 

0  =  fc, .  4,2  +  P.  12  (rot.  r.  6)  J 
a-3=r  85700.  ' 

For  $,  the  intersection  £  of  rc3  and  z3  is  to  the  left  of  the  sus- 
pended weight,  and  the  symbql  reversed. 

0  =  y,  •.  7,25  +  P.  2,6  (rot.  r .  I) } 
y,  =  — 10758, 

which  would  be  =  0  when  the  intersection  is  in  the  vertical  line  of 
the  suspended  weight,  as  the  lines  oe  and  pc  in  Fig,  A  indicate. 
For  the  verticals,  V,  we  have  from  Fig.  D, 

0  =  —  Fi  -.  10,1  —  P.  6,1  (rot.  r .  m)  ; 

F,  =  18118 ; 
0  =  —  F2 .  26,5  —  P.  18,5  (rot.  r.n)\ 

V,  =  —  20943; 
E.]     The  results  combined  in  Fig.  E.* 

II.    SEMI-GIRDERS  LOADED  AT  EACH  APEX. 

In  Fig.  25  is  occasionally  explained  how  to  compute  the  stress  irt 
diagonals,  as  there  is  no  intersection  of  joining  flanges,  x  and  z,  and 
fcs  in  the  case  here  considered  the  diagonals  receive  at  each  loaded 


*  For  most  purpose?  the  above  will  be  sufficient.     In  Giynn's  rudimentary 
'treatise  on  the  Construction  of  Crnncs  we  find  valuable  and  complete  drawings 


46  THE  THEORY   OF  STRAINS. 

PI       1^1    aPex  an  increment  of  strain,  prior  to  the  calculation 
pj       gQj    may  be  given  the  general  thesis  that  the  strain  in  two 
diagonals  whose  intersection  is  at  an  unloaded  point  is 
the  same  in  numerical,  value,  but  of  opposite  character.     (Fig.  80.) 

(See  IV.  General  Remarks.) 

The  strain  in  diagonals,  meeting  at  a  loaded  point,  is  in  numeri- 
cal value  different. 

The  strain  in  flanges  increases  from  apex  to  apex  in  geometrical 
progression. 

£*  -.        In  Fig.  81  suppose  the  angle  <p  of  diagonals  with  a  hori- 
zontal line  =  45°,  so,  also,  angle  oc  =  45°;  and  by  the  table, 
sec  oc  =  1,414. 

When,  again,  in  the  axis  x  =  GO  a  point  of  rotation,  o,  is  sup- 
posed, we  have  per  example  for  diagonal,  i/5. 

/  W\ 

0  =  -f  ?/5  .  x  sin  <p  -f  I    W+  W  +  -—  .  x  (rot.  r  .  o), 


where  y-3  .  sin  <p  is  the  vertical  component  of  y5,  or  —  a.b  —  ?.&  in 
the  parallelogram  of  forces  (Fig.  81),  presenting  by  y5  the  resulting 
strain  or  diagonal.  Divided  by  x,  it  follows  : 


\ 
and  as  <p  =  45°,  and  sin  45°  =  0,707, 

0  =  +  y6.  0,707  +  I-.TF, 
or  y,  =  —  3,535  W. 

The  same  results  from  —  f  W.  sec  oc  ,  or  —  f  .  W  .  1,414,  which  13 
also  =  —  3,535  W. 

82.]        *n  ^ie  sarue  manner  *n  Fig.  82  for  y2,  the  direction  of 
which,  when  separated  by  a  cut,  st,  is  reversed  to  the  weight. 
(See  Figs.  22  and  23.) 

W 

-  y,  .  sin  45  +  —  (rot.  o)  ; 


0,707  ' 
and  for  the  other  diagonals, 

0^  +  2/3.0,707+  TF  +  —  , 


ib«---, 

0,707  ^        r  0.707  ' 


'  •*» 


*   I 


SEMI-GIRDERS    LOADED    AT    EACH    APEX.  47 


TT  fir 


f-TF 
or  yr  =  —  r 


0,707  0,707 

83.]     For  the  strain  in  flanges  we  have  from  Fig.  83, 

W 
0==  —  Xi .  I .  cot.  QC  -\ •  I  (rot.  r .  1) ; 

%W.l  1 

KI  =  — *      — »      or  as =  tang  oc  , 

I.  cotg  QC  cotg  oc 

x1  =  ±W.  tang  oc  ; 

and  when     W=  10  tons,  <  y  =  60°;  therefore  <  oc  =  30°, 
and  tang  30°  =  0,577  ("  Example  Stoney") ; 

Xl  =  5 .  X  0,577  =  +  2,9  tons ; 
or  when,  for  an  easier  understanding,  in  Fig.  83, 
I .  cotg  oc  =  h, 

84.]    we  have  for  x2  in  Fig.  84, 

W 

Q  =  —  x.1.h+W.l+  —  -Zl  (rot.  r.  3) ; 

JO 

IW.l 

n-   —  A . 


k 

and  as  =  tang  oc  =  0,577, 

h 
for  our  example, 

x,  =  f .  W.  0,577  =  |  X  10  X  0,577  =  14,5  tons. 
85.]     So  for  xs  in  Fig.  85, 

0  =  —  a,,  h  +  W.  I  +  TF.  3  I  +  —  •  5 1  (rot.  r .  5)  ; 

Zt 

x.  =  LS.  |p.  o,577  =  37,3  tons, 
and  so  further. 

86.]     For  the  strain,  2,  in  the  lower  flanges, 

W 

0  =  +  2l .  h  +  —  •  2  /  (rot.  r .  2  in  Fig.  86) ; 
<u 

•i  =  —  Tr4  =  —  10  X  0,577  =  —  5,7  tons. 
h 


48  THE   THEORY    OP   STRAINS 

87.]    Fig.  87  gives 


r.4); 


Jh==_4jp.==_  4X  10  X  0,577  =  —  23,2  tons, 
h 

In  the  same  way  for  zq, 


23  —  —  9  W-  -  =  —  52  tons. 
/i. 

In  case  the  load  should  be  connected  to  the  lower  apexes 
•     (Fig.  88),  the  equation  of  equilibrium,  per  example  for  s3l 

would  be 

W 

—  -51  (rot,  r  .m)  ; 


Z3=-  if.  W-  r  =  —  37,3  tons  ; 

/I/ 

L  e.,  the  strain  is  the  same  as  in  the  flange  of  the  reversed  figure, 
but  of  opposite  character.  So  also  is  the  strain  the  same  for  the 
other  flanges,  (See  Fig.  88.) 

Remark.  —  In  the  given  example  the  strains  are  determined  with- 
out a  certain  length  for  h  or  L  This  is  easily  explained  by  the 
relation  which  the  angle  <p  or  <x.  bears  to  h  and  I,  as  by  the  exten- 
sion of  one,  the  other  will  increase  in  the  same  ratio. 

[Plates  13,  14  and  15  —  embracing  Figs.  71  to  88.] 


D.   GIRDERS  WITH  PARALLEL  TOP  AND  BOTTOM 
FLANGES. 

( Calculated  for  a  Permanent  Load.) 
I.   STRAIN  IN  DIAGONALS  AND   VERTICALS. 

The  calculation  is  very  similar  to  the  preceding.  Provided, 
again,  the  load  to  be  connected  to  the  upper  or  lower  apexes  for 
the  application  of  the  Method  of  Moments,  we  now  consider  as  a 
special  force  the  reaction  of  the  supports  toward  the  system. 

D,  may  represent  one-half  the  weight  of  loaded  truss  or  the 


STRAIN   IN    DIAGONALS  ASfD   VERTICALS.  49 

pressure  upon  each  support  (prop),  and  diminished  by  the  par- 
tition of  load  on  this  place  directly  sustained.  (Comp.  Fig.  21.) 
The  reactive  force  of  support  wanted  for  our  calculation  will  be 
signified  by  D. 

To  compute  D,  we  refer  to  Fig.  4  in  Sect.  I.,  and  define  first  for 
the  following  example  its  numerical  value : 

Plate  16,1  Through  bridge  (over-grade  bridge),  between  supports, 
Fig.  89.J  48  feet; 

8  panels,  each  6  feet  =  I ; 

depth  of  truss  =  6  feet  =  h,  from  centre  to  centre  of  top  and 

bottom  chords ; 

the  weight  of  structure  =  3000  Ibs.,  and  the  load  =  16000  Ibs. ; 
gives  a  permanent  load  =  18000  Ibs.  per  panel ; 
rolling  load  =  0.     (See  Sect.  II.) 

For  the  distribution  of  load,  see  Fig.  89. 

Remark. — The  strange  impression  which  the  arrangement  of 
diagonals,  unsymmetrical  toward  the  centre,  may  first  produce,  will 
soon  disappear  after  observation  of  the  advantages  for  transforma- 
tion upon  succeeding  systems. 

Whole  pressure  of  truss  upon  supports  =  8  X  18000  =  144000  Ibs. ; 
A  =  18000(i  +  f  +  f  +  t  +  *  +  i  +  l)  +  9000XI-72000  Ibs.; 
D  =  18000  (^+1  +  1  +  1+1  +  1  +  1)  =  63000  Ibs. 

The  strain  in  the  post,  V0,  is  0,  because  x^  =  0, 
and  F8  =  —  63000.* 

q^  -,        Excepting  the  vertical  component  of  yt  (i.  e.,  yi  sin  $0),  for 
a  section  (Fig.  90),  only  D  =  63000  Ibs.  is  a  second  vertical 
force. 

Both  turn  to  the  left  around  o  in  the  axis  x0 ;  therefore,  their 
symbol,  — .  (Comp.  Fig.  22.) 

*  For  a  deck-bridge  (under-grade  bridge) — i.  e.,  when  the  upper  apexes  ar« 
loaded — would  be 

V0  =  —  9000,        and  F8  =  —  72000. 

The  angle  of  diagonals  with  a  horizontal  line  will  be  45°, 
5  D 


